Brian Bi

## Section 2.6. Isomorphisms

Exercise 2.6.2 The homomorphism $$\varphi$$ is completely determined by the image of 1, since 1 generates the domain. If $$\varphi(1) = 0$$, then $$\varphi$$ is neither surjective nor injective. If $$\varphi(1) = \pm 1$$, then $$\varphi(n) = \pm n$$, and $$\varphi$$ is both surjective and injective (hence, an isomorphism). If $$\varphi(1) = k$$ where $$k \neq \pm 1$$, then $$\varphi(n) = kn$$ and $$\varphi$$ is injective, but not surjective as 1 is not in the image.

Exercise 2.6.4 $$ba = b(ab)b^{-1}$$, the conjugate of $$ab$$ by $$b$$.

Exercise 2.6.5 The eigenvalues of $$B$$ are 2 and 3. Since $$B$$ is a 2×2 matrix, the multiplicity of each eigenvalue is exactly 1, therefore $$A$$ is the Jordan form of $$B$$ (up to the order of diagonal elements). Therefore $$A$$ and $$B$$ are similar, which is the same as saying that they are conjugate in $$GL_2(\mathbb{R})$$.

Exercise 2.6.6 Denote the two given matrices by $$A, B$$. They are conjugate by the matrix $$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$, so they are conjugates in $$GL_2(\mathbb{R})$$. In general, $$M = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$$, then solving the equation $$BM = MA$$ yields the constraints $$p = 0, q = r$$, implying $$\det M = -q^2$$, which is not possible when $$\det M = 1$$ and $$q$$ is real. Therefore $$A$$ and $$B$$ are not conjugate elements in $$SL_2(\mathbb{R})$$.

Exercise 2.6.7 Since $$1 \in H$$, it follows that $$g(1)g^{-1} = 1$$ is in $$gHg^{-1}$$. For all $$k \in gHg^{-1}$$, there eixsts some $$h \in H$$ with $$k = ghg^{-1}$$, whence $$k^{-1} = gh^{-1}g^{-1}$$, which lies in $$gHg^{-1}$$ since $$h^{-1} \in H$$. Finally, for all $$k_1, k_2 \in gHg^{-1}$$, there exist some $$h_1, h_2 \in H$$ with $$k_1 = gh_1 g^{-1}, k_2 = gh_2 g^{-1}$$, implying $$k_1 k_2 = gh_1 g^{-1} g h_2g^{-1} = g(h_1 h_2)g^{-1} \in gHg^{-1}$$ since $$h_1 h_2 \in H$$. Therefore $$gHg^{-1}$$ is a subgroup of $$G$$.

Exercise 2.6.8 For all $$A, B \in GL_n(\mathbb{R})$$, we have $$\varphi(AB) = ((AB)^t)^{-1} = (B^t A^t)^{-1} = (A^t)^{-1} (B^t)^{-1} = \varphi(A)\varphi(B)$$. Therefore $$\varphi$$ is an endomorphism of $$GL_n(\mathbb{R})$$. It is easy to see that the kernel of $$\varphi$$ is trivial, making $$\varphi$$ an automorphism.

Exercise 2.6.9 We claim that the map $$\varphi$$ from $$G$$ to $$G^\circ$$ given by $$\varphi(g) = g^{-1}$$ is an isomorphism. (We established in Exercise 2.2.6 that $$G^\circ$$ is indeed a group.) Let $$a, b \in G$$. Then $$\varphi(ab) = (ab)^{-1} = b^{-1} a^{-1} = a^{-1} * b^{-1} = \varphi(a) * \varphi(b)$$. Therefore $$\varphi$$ is a homomorphism. Its kernel is trivial, so it is an isomorphism.

Exercise 2.6.10

1. Let $$\varphi : C_{10} \to C_{10}$$ be an automorphism. Since $$C_{10}$$ is cyclic, the homomorphism $$\varphi$$ is entirely determined by $$\varphi(1)$$, since $$\varphi(2) = \varphi(1) + \varphi(1)$$ and so on; in general, every valid endomorphism must take the form $$\varphi(n) = kn$$. For such $$\varphi$$, we have $$\varphi(a+b) = k(a+b) = ka + kb = \varphi(a) + \varphi(b)$$ for all $$a, b$$, so $$\varphi$$ preserves the group operation. Therefore the set of such $$\varphi$$ is exactly the set of endomorphisms of $$C_{10}$$. Note that $$\varphi$$ can only take on values that are divisible by $$\gcd(k, 10)$$, so $$k$$ must be congruent to 1, 3, 7, or 9 in order for $$\varphi$$ to be an automorphism. As proven in the solution to Exercise 2.4.6, the elements 1, 3, 7, and 9 generate $$C_{10}$$, so $$\varphi(n) = kn$$ for $$k \in \{1, 3, 7, 9\}$$ maps a generator to a generator, implying that it is surjective, and hence, in this case, also injective, therefore an automorphism. Therefore, the automorphisms of $$C_{10}$$ are $$n \mapsto n, n \mapsto 3n, n \mapsto 7n$$, and $$n \mapsto 9n$$.
2. An endomorphism $$\varphi : S_3 \to S_3$$ is completely determined by the images of the generators $$x, y$$; there will be exactly one valid extension to all of $$S_3$$ provided that $$\varphi(x)$$ and $$\varphi(y)$$ are chosen so that $$\varphi(y)\varphi(x) = \varphi(x)^2 \varphi(y) \label{eqn:e2}$$ that is, $$\varphi$$ also preserves the defining relation of $$S_3$$. If $$\varphi$$ is additionally an automorphism, it must preserve the orders of elements. Referring to the table from Exercise 2.2.1, we see that $$x$$ and $$x^2$$ have order 3 while $$x^j y$$ has order 2 for $$j \in \{0, 1, 2\}$$. Therefore each valid automorphism must take the form $$\varphi_{ij}$$ which satisfies $$\varphi_{ij}(x) = x^i, \varphi_{ij}(y) = x^j y$$, with $$i \in \{1, 2\}$$ and $$j \in \{0, 1, 2\}$$. Then \begin{align} \varphi(y)\varphi(x) &= x^j y x^i \label{eqn:e1} \\ \varphi(x)^2\varphi(y) &= x^{2i} x^j y \end{align} Since $$yx = x^2 y$$, we can move the $$y$$ factor $$i$$ positions to the right in $$(\ref{eqn:e1})$$ to obtain $$\varphi(y)\varphi(x) = x^{2i+j} y$$; therefore $$(\ref{eqn:e2})$$ is always satisfied, so the 6 different possible maps $$\varphi_{ij}$$ are all valid endomorphisms of $$S_3$$.

The general formula for $$\varphi_{ij}$$ is, of course, $$\varphi_{ij}(x^m y^n) = \varphi_{ij}(x^m) \varphi_{ij}(y^n) = x^{im} (x^j y)^n$$ Suppose this is equal to 1. Now $$x^j y$$ has order 2 so a factor of $$y$$ will be left in the result unless $$n$$ is even. In the case where $$n$$ is even, $$\varphi_{ij}(x^m y^n) = x^{im}$$, so $$im$$ must be divisible by 3, and since $$i$$ is either 1 or 2, this implies $$m$$ is divisible by 3. Therefore, if $$\varphi_{ij}(g) = 1$$ then $$g = 1$$, that is, each of the 6 $$\varphi_{ij}$$'s is an automorphism of $$S_3$$, and there are no other automorphisms.

Exercise 2.6.11 Since $$\{1, a\}$$ is a normal subgroup, for all $$g \in G$$, $$gag^{-1} \in \{1, a\}$$. But $$gag^{-1}$$ cannot be equal to 1 unless $$a = 1$$, therefore $$gag^{-1} = a$$ for all $$g \in G$$, that is, $$a$$ is central.